Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in PYTHON3):

Solution:
n=int(input())
p=list(map(float,input().split()))
# 最初の ii 枚のコインを投げたときに、表が jj 枚となる確率
dp=[[0.0]*(n+1) for i in range(n+1)]
dp[0][0]=1.0
for i in range(n):
  for j in range(n):
    dp[i+1][j+1]+=dp[i][j]*p[i]
    dp[i+1][j]+=dp[i][j]*(1-p[i])
    
ans=0.0
for i in range((n+1)//2,n+1):
  ans+=dp[n][i]
print(ans)